Question

A small sphere with mass $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{\hspace{0.33em}}\mathbf{kg}$and charge $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}\mathbf{C}$hangs from a thread near a very large, charged insulating sheet (Fig.P22.33). The charge density on the surface of the sheet is uniform and equal to $\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{}\mathbf{C}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$. Find the angle of the thread.

Short answer

The angle of the thread is $10.2°$

Step-by-step solution

Define free body diagram and given data

Free body diagram,

Given,

$$\begin{gathered} \mathrm{m}=4\times {10}^{-6}\mathrm{kg} \\ \mathrm{\sigma }=2.50\times {10}^{-9}\mathrm{C}\times {\mathrm{m}}^{-2} \\ \mathrm{q}=5.0\times {10}^{-8}\mathrm{C} \end{gathered}$$

Simplify the direction of force

The ball is in equilibrium so the net force in each direction is equal to zero so the force indirection is given by

$\mathrm{Tsin}\left(\mathrm{\theta }\right)={\mathrm{F}}_{\mathrm{e}}\hspace{0.33em}......\left(1\right)$

Also, the force in y direction is

$\mathrm{Tcos}\left(\mathrm{\theta }\right)=\mathrm{mg}....\left(2\right)$

Dividing (1) over (2) to get

$\tan \left(\mathrm{\theta }\right)=\frac{{\mathrm{F}}_{\mathrm{e}}}{\mathrm{mg}}\hspace{0.33em}\mathrm{Þ}\left(3\right)$

The electric force is given by

$$\begin{gathered} {\mathrm{F}}_{\mathrm{e}}={\mathrm{E}}_{\mathrm{q}} \\ =\frac{\mathrm{q\sigma }}{2{\displaystyle \stackrel{\text{'}}{{\mathrm{o}}_0}}} \\ =\frac{2.5\times {10}^{-9}\times 5\times {10}^{-8}}{2\times 8.85\times {10}^{-12}} \\ =7.06\times {10}^{-6}\mathrm{N} \end{gathered}$$

Also, the gravitational force is given by

$$\begin{gathered} \mathrm{mg}=4\times {10}^{-6}\times 9.8 \\ =3.9\times {10}^{-5}\mathrm{N} \end{gathered}$$

Substitution in (3) to get

$$\begin{gathered} \tan \left(\theta \right)=\frac{7.06\times {10}^{-6}}{3.9\times {10}^{-5}} \\ =0.18 \\ \theta =10.2° \end{gathered}$$

Hence, the angle of the thread is $10.2°$